Let A = left| begin{array}{cc} 2 & e^{i pi} | vedclass

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(A) First,we calculate the values of $A$,$C$,and $D$: $A = \left| \begin{array}{cc} 2 & \cos \pi + i \sin \pi \ -1 & (i^4)^{503} \end{array} ight|

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(A) First,we calculate the values of $A$,$C$,and $D$: $A = \left| \begin{array}{cc} 2 & \cos \pi + i \sin \pi \ -1 & (i^4)^{503} \end{array} ight| = \left| \begin{array}{cc} 2 & -1 \ -1 & 1 \end{array} ight| = (2)(1) - (-1)(-1) = 2 - 1 = 1$. $C = \left. -\frac{1}{x^2} ight|_{x=1} = -1$. $D = [\ln x]_{e^2}^{1} = \ln 1 - \ln e^2 = 0 - 2 = -2$. Substituting these into the equation $Ax^3 + Bx^2 + Cx - D = 0$,we get: $1x^3 + Bx^2 - 1x - (-2) = 0 \Rightarrow x^3 + Bx^2 - x + 2 = 0$. Let the roots be $\alpha, \beta, \gamma$. Given $\alpha + \beta = 0$. From Vieta's formulas,$\alpha + \beta + \gamma = -B$,so $0 + \gamma = -B$,which means $\gamma = -B$. Since $\gamma$ is a root,it must satisfy the equation: $(-B)^3 + B(-B)^2 - (-B) + 2 = 0$ $-B^3 + B^3 + B + 2 = 0$ $B + 2 = 0 \Rightarrow B = -2$. Wait,re-evaluating the equation sign: The equation is $Ax^3 + Bx^2 + Cx - D = 0$. With $D = -2$,the equation is $x^3 + Bx^2 - x - (-2) = 0$,which is $x^3 + Bx^2 - x + 2 = 0$. Substituting $\gamma = -B$: $(-B)^3 + B(-B)^2 - (-B) + 2 = 0 -B^3 + B^3 + B + 2 = 0 B = -2$.

Let $A = \left| \begin{array}{cc} 2 & e^{i \pi} \\ -1 & i^{2012} \end{array} \right|$,$C = \left. \frac{d}{dx} \left( \frac{1}{x} \right) \right|_{x=1}$,and $D = \int_{e^2}^{1} \frac{dx}{x}$. If the sum of two roots of the equation $Ax^3 + Bx^2 + Cx - D = 0$ is equal to zero,then $B$ is equal to:

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